Discussion and Proof:

If x is a perfect square in $\mathbf{Z}^*_p$ and g is a generator of $\mathbf{Z}^*_p$, x can be written as $g^{2i} \mbox{ mod } p \mbox{ where } 1 \leq 2i \leq p-1$. As a result, $\mathbf{J}_p(x) = g^{\frac{2i(p-1)}{2}} = g^{i(p-1)} = 1 \mbox{ mod }p$

If x is not a perfect square in $\mathbf{Z}^*_p$ and g is a generator of $\mathbf{Z}^*_p$, x can be written as $g^{i} \mbox{ mod } p \mbox{ where } 1 \leq i \leq p-1$. Since i is odd, $\frac{i(p-1)}{2}$ is not an integer and $\mathbf{J}_p(x) = g^{\frac{i(p-1)}{2}} = \sqrt{1} = \pm{1}$. However, this does not quite match the definition. To continue the proof, we have to prove thatx is a perfect square in $\mathbf{Z}^*_p$ if and only if $x^{\frac{p-1}{2}} = 1 \mbox{ mod } p$:
$\Rightarrow$ Assume x is a perfect square in $\mathbf{Z}^*_p$, x can be written as $g^{2i} \mbox{ mod } p \mbox{ where } 1 \leq 2i \leq p-1$ and $\mathbf{J}_p(x) = g^{\frac{2i(p-1)}{2}} = g^{i(p-1)} = 1 \mbox{ mod }p$
$\Leftarrow$ Assume $x^{\frac{p-1}{2}} = 1 \mbox{ mod } p$. Since $x \in \mathbf{Z}^*_p$, x can be written as $g^{i} \mbox{ mod } p$ where $1 \leq i \leq p-1$. Since g is a generator of $\mathbf{Z}^*_p$, g has the order $(p-1)$. Only $g^{p-1} \equiv 1 \mbox{ mod } p$ and $g^i \not\equiv 1 \mbox{ for } i < p-1$. As a result, $(p-1)$ must divide $\frac{i(p-1)}{2}$ and so i is even and x is a perfect square in $\mathbf{Z}^*_p$
As a result, $\mathbf{J}_p(x) \neq 1$ if x is not a perfect square in $\mathbf{Z}^*_p$ and so $\mathbf{J}_p(x) = -1$